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Data Structures and Libraries problem Solution

Data Structures and LibrariesData Structures and Libraries Problems  are more interesting to solve.These problem are are not easy but solvable.Some of the problems type Like: Basic Data Structures, Sorting-related problems, Static array, vector, bitset, Direct Addressing Table,STL stack,queue,STL priority_queue are the Data Structures problem With Built-in Libraries and also solvable.

Generally most Data Structures and Libraries Problems  are easy and can be solved by previous

knowledge & with help of internet.  But be careful about time, space and special criteria when solving them.Here are the some data structures and libraries problem given, UVa-299 -Train Swapping,UVa-146-ID Codes, UVa-673-Parentheses Balance,UVa-11340-Newspaper.

Some Data Structures and Libraries UVa Problems Solution are given below:

UVa: 146 – ID Codes

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main () {
char c[65];
while (scanf("%s",c)==1 && c[0]!='#') {
if (next_permutation (c,c+strlen(c)))
printf("%s\n",c);
else
printf("No Successor\n");
}
return 0;
}

UVa:  299 – Train Swapping

#include <stdio.h>

int main (){
long int n,a[55],x,c,i,j,t;

scanf("%ld",&n);

while (n--){

scanf("%ld",&x);
for (i=1; i<=x; i++)
scanf("%ld",&a[i]);

c=0;
for (i=1; i<=x; i++)
for (j=i+1; j<=x; j++)
if (a[i]>a[j]){
t=a [i];
a[i]=a[j];
a[j]=t;
c++;
}

printf("Optimal train swapping takes %ld swaps.\n", c);
}
return 0;
}

UVa:  673 – Parentheses Balance

#include<stdio.h>
#include<string.h>
int main(){
int i,n,top;
char a[150],b[150],ch;
scanf("%d%c",&n,&ch);
while(n-->0){
top=0;
gets(a);
for(i=0;i<strlen(a);i++){
if(a[i]=='(')
b[top++]=a[i];
else if(a[i]=='[')
b[top++]=a[i];
else if(a[i]==')'&&top!=0&&b[top-1]=='(')
top--;
else if(a[i]==']'&&top!=0&&b[top-1]=='[')
top--;
else
break;
}

if(top==0 && i==strlen(a))
printf("Yes\n");
else
printf("No\n");
}
return 0;
}

UVa:  11340 – Newspaper

#include <stdio.h>
#include <string.h>

int n,p,v;
unsigned long long int sum;
char c,k,po[10001];

int main()
{

scanf("%ld",&n);

while(n--)
{
sum=0;
int ch[300]={0};
scanf("%d%c",&p,&c);

while(p--)
{
scanf("%c %d%c",&k,&v,&c);
ch[k]=v;
}

scanf("%d%c",&p,&c);
while(p--)
{
gets(po);
for(v=0; v<strlen(po); v++)
if(ch[po[v]]) sum+=ch[po[v]];
}
printf("%.2f$\n",sum/100.00);
}

return 0;
}

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